Prove that ${\tan ^{ – 1}}\frac{1}{4} + {\tan ^{ – 1}}\frac{2}{9} = {\sin ^{ – 1}}\frac{1}{{\sqrt 5 }}$.
[NCERT,Exemplar.2.3,Q.16,Page.36]

We have, ${\tan ^{ – 1}}\frac{1}{4} + {\tan ^{ – 1}}\frac{2}{9} = {\sin ^{ – 1}}\frac{1}{{\sqrt 5 }}$

….(i)
Let ${\tan ^{ – 1}}\frac{1}{4} = x$
$ \Rightarrow $$\tan x = \frac{1}{4}$

$ \Rightarrow $${\tan ^2}x = \frac{1}{{16}}$
$ \Rightarrow $${\sec ^2}x – 1 = \frac{1}{{16}}$

$ \Rightarrow $${\sec ^2}x = 1 + \frac{1}{{16}} = \frac{{17}}{{16}}$

$ \Rightarrow $$\frac{1}{{{{\cos }^2}x}} = \frac{{17}}{{16}}$

$ \Rightarrow $${\cos ^2}x = \frac{{16}}{{17}}$
$ \Rightarrow $$\cos x = \frac{4}{{\sqrt {17} }}$

$ \Rightarrow $${\sin ^2}x = 1 – {\cos ^2}x = 1 – \frac{{16}}{{17}} = \frac{1}{{17}}$

$ \Rightarrow $$\sin x = \frac{1}{{\sqrt {17} }}$

…..(ii)
Again, let ${\tan ^{ – 1}}\frac{2}{9} = y$
$ \Rightarrow $$\tan y = \frac{2}{9} \Rightarrow {\tan ^2}y = \frac{4}{{81}}$

$ \Rightarrow $${\sec ^2}y – 1 = \frac{4}{{81}}$
$ \Rightarrow $${\sec ^2}y = \frac{4}{{81}} + 1 = \frac{{85}}{{81}}$

$ \Rightarrow $${\cos ^2}y = \frac{{81}}{{85}} \Rightarrow \cos y = \frac{9}{{\sqrt {85} }}$

$ \Rightarrow $${\sin ^2}y = 1 – {\cos ^2}y = 1 – \frac{{81}}{{85}} = \frac{4}{{85}}$

$ \Rightarrow $$\sin y = \frac{2}{{\sqrt {85} }}$

……(iii)
We know that,
$\sin (x + y) = \sin x \cdot \cos y + \cos x \cdot \sin y$

$ = \frac{1}{{\sqrt {17} }} \cdot \frac{9}{{\sqrt {85} }} + \frac{4}{{\sqrt {17} }} \cdot \frac{2}{{\sqrt {85} }}$

$ = \frac{{17}}{{\sqrt {17} \cdot \sqrt {85} }} = \frac{{\sqrt {17} }}{{\sqrt {17} \cdot \sqrt 5 }} = \frac{1}{{\sqrt 5 }}$

$ \Rightarrow $$(x + y) = {\sin ^{ – 1}}\frac{1}{{\sqrt 5 }}$

$ \Rightarrow $${\tan ^{ – 1}}\frac{1}{4} + {\tan ^{ – 1}}\frac{2}{9} = {\sin ^{ – 1}}\frac{1}{{\sqrt 5 }}$ Hence proved.

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